tisdag 14 april 2015

Unphysical Schwarzschild vs Physical Model for Radiative Transfer

The basic model for radiative heat transfer in a horisontal slab atmosphere, with vertical coordinate $x$ with $x=0$ at the Earth surface and $x=X$ at the top of the atmosphere, was formulated by Schwarzschild in 1906 as a two-stream  gross-flow model in terms of a gross upward heat flux $F^+(x)$ and a gross downward heat flux  $F^-(x)$ satisfying the following advection-absorption equations for $0\lt x\lt X$:
• $\frac{dF^+}{dx} + F^+ = Q$               (1)
• $-\frac{dF^-}{dx} + F^- = Q$               (2)
where $Q(x) =\sigma T(x)^4$ is supposed to express Stefan-Boltzmann's law with $T(x)$ the temperature at $x$ and $\sigma$ Stefan-Boltzmann's constant, and $x$ serves as an optical coordinate normalizing absorption. The atmosphere is supposed to be heated from below at $x=0$ by a heat source $H$, and the heat is radiatively transported to the top of the atmosphere from where it is radiated into outer space at 0 K. Conservation of heat energy gives the additional equation
• $F^+-F^- = H$,                                      (3)
from which follows by adding/subtracting (2) from (1) that $F^+ + F^-=2Q$ and $\frac{d(F^++F^-)}{dx}=-H$ and thus:
• $2Q(x) = H(X-x)+H$,                          (4)
• $F^+ =\frac{H}{2}(X-x)+H$
• $F^-=\frac{H}{2}(X-x)$
which determines the temperature profile $T(x)$. We note that Schwarzschild's model with linear $Q(x)$, is very simplistic. Only a model with $Q(x)$ constant could be more simplistic.

Schwarzschild's equations (1-2) are supposed to express conservation of upward and downward heat fluxes through a thin atmospheric layer radiating both upward and downward according to Stefan-Boltzmann in the form $Q(x) =\sigma T(x)^4$.

Schwarzschild's two-stream model is unphysical in the sense that the gross downward flux $F^-$ is directed against the temperature gradient and thus violates the 2nd law of thermodynamics. Further, the two sided radiation up/down according to $Q(x) =\sigma T(x)^4$ in (1-2) is also unphysical, because  $Q(x) =\sigma T(x)^4$ is the blackbody radiation into a background at 0 K, which is not the case. Scwarzschild's model is thus doubly unphysical and we shall now see that the unphysical aspects do not cancel to give a physical model.

Let us thus compare Schwarschild's unphysical two-stream gross-flow model with a one-stream net-flow model based on Stefan-Boltzmann's law in its correct physical form
• $Q=\sigma (T_2^4-T_1^4)$                     (5)
as the net radiative heat transfer radiative between two blackbodies of temperature $T_1$ and $T_2$ with $T_2>T_1$, in accordance with the 2nd law with heat transfer from warm to cold (but not the other way).

We then start from the following balance equation expressing that the total inflow of heat at a thin layer at level $0\lt x\lt X$ from layers at levels $y$ with $0\le y\lt x$, equals the total outflow into levels $x\lt y\le X$:
• $\int_0^x\sigma (T(y)^4-T(x)^4)\exp(-\alpha (x-y))dy = \int_x^X\sigma (T(x)^4-T(y)^4)\exp(-\alpha (y-x))dy$
• $+H\exp(-\alpha x)$                                                         (6)
with the exponential factor $\exp(-\alpha\vert x-y\vert )$ accounting for the absorption between levels $x$ and $y$ with a non-negative absorption coefficient $\alpha$, and the term $H\exp(-\alpha x)$ accounts for the effect of the heat source $H$ at $x=0$, and $\sigma T(X)^4 = H$.

Reshuffling terms in (6), we obtain the following integral equation in $Q(x)=\sigma T(x)^4$ for $0\le x\le X$
• $\int_0^xQ(y)\exp(-\alpha (x-y))dy+ \int_x^XQ(y)\exp(-\alpha (y-x))dy$
• $= Q(x)(\int_0^x\exp(-\alpha (x-y))dy+ \int_x^X\exp(-\alpha (y-x))dy)+H\exp(-\alpha x)$,
which can be written
•  $Q(x)-c(x)\int_0^XQ(y)\exp(-\alpha\vert x-y\vert )dy =-c(x)H\exp(-\alpha x)$   (7)
for $0\le x\le X$ with $c(x) = 1/(2-\exp(-\alpha x)-\exp(-\alpha (X-x)),$ which together with the outflow condition $Q(1) = H$ determines $Q(x)$ uniquely as the solution of a Fredholm integral equation of the 2nd kind.

To compare with the two-stream model, let us formally split Stefan-Boltzmann's equation (5) algebraically into the (unphysical) form
• $Q=\sigma T_2^4-\sigma T_1^4$                   (8)
and rewrite (6) accordingly collecting positive terms and writing $Q(x)=\sigma T(x)^4$ as above, into
• $A+B = C+D+E$                                                  (9)
• $A(x)= \int_0^xQ(y)\exp(-(x-y))dy$
• $B(x)= \int_x^XQ(y)\exp(-(y-x))dy$
• $C(x)=Q(x)(1-\exp(-x))$
• $D(x)=Q(x)(1-\exp(-(X-x))$
• $E(x) = H\exp(-\alpha x)$.
To compare the models we observe that by the definitions of $A$ and $B$:
• $\frac{dA}{dx}+A = Q$
• $-\frac{dB}{dx}+B = Q$
which shows that $A$ corresponds to $F^+$ and $B$ to $F^-$.  We recall that
• $F^+ + F^-=2Q$
which we we compare with (9):
• $A+B=C+D+E=Q(2-\exp(-x)-\exp(X-x))+H\exp(-\alpha x)$,
to conclude that the two models are not identical, and so the models give different temperature distributions.

The lesson is that radiative heat transfer should better be modeled using the physical one-stream net-flow correct form of Stefan-Boltzmann's law (5). Using the unphysical two-stream gross-flow form (1-2) can lead to unphysical results.

It is possible that an unphysical gross-flow model can give a physically meaningful result by happy cancellation of unphysical gross-flow aspects. But to rely on an unphysical model to derive conclusions about real physics is risky because the happy cancellation may not be granted, in particular not if the question concerns the effect of perturbations, as is the case in global warming modeling.

The unphysical aspects of Schwarzschild's model are:
1. The downward flux $F^-$ violates the 2nd law.
2. The downward flux $F^-$ generates unphysical absorption in (2).
3. The basic equations (1) and (2) do not represent correct physics.
Despite these shortcomings, Schwarzschild's model has come to serve as the basic model of atmospheric radiation supporting CO2 alarmism.  The unphysical of nature of this basic model gives one reason to view also CO2 alarmism to be unphysical.

In a following post we will solve the integral equation (7) and compare with Schwarzschild's solution. To start with we note that for $\alpha$ large (7) approaches $\frac{d^2Q}{dx^2}=0$ resulting in a linear $Q(x)$ as in Schwarzschilds model. More generally, the physical model (7) is close to Schwrzschilds model in the trivial cases of a nearly opaque and transparent atmosphere, but not so in the more relevant case in between.

The integral equation (7) can by differentiation be turned into a (diffusion-advection-absorption) ordinary linear differential equation.

14 kommentarer:

1. I think this must be the final nail in the AGW "back radiation" dogma, and thus there is no credibility left. However, it has never been about science, but politics, power and "redistribution of wealth" unfortunately from poor people to the rich elite and AGW activists.

2. Why does the downward radiative flux violate the 2nd law.
It is not a heat flux.

3. The downward flux F− violates the 2nd law.

This statement of Mr. Johnson is wrong. Downward Flux is not heat.

4. Your Statement #1 is wrong. Downward flux is not heat.

5. What kind of flux is it then? In what units is it measured?

6. It is an energy flux since it doesn't depend on temperature differences.

If you sum up all energy fluxes over a control surface you get the heat flux.

It is measured in [W/m^2] or [W] if you integrate it over a control surface.

7. 1) You are using Stefan-Boltzmanns law for a perfect black body (i.e. a body that absorbs/emits radiation on all wavelengths), to get a more realistic model you must include the emissitivity of each body < 1.
2) You state: "Schwarzschild's two-stream model is unphysical in the sense that the gross downward flux F− is directed against the temperature gradient and thus violates the 2nd law of thermodynamics."
Physics does not state that there will be a 'gross downward flux F-', as mentioned the emissitivity will be less than 1, i.e the upper body will always absorb/emit less than 1. You can only argue this if you somehow increase emisitivity to > 1.
3) Stefan-Boltzmann states that absorption/emission ONLY depends on the temperature of the body itself. You find no dependencies on the surroundings.
This is also in accordance with what Max Planck states in his lectures:
"The capacity for emission and the capacity for absorption of an element of a body depend only upon its own condition and not upon that of the surrounding elements. If, therefore, as we shall assume in what follows, the state of the body varies only with the temperature, then the capacity for emission and the capacity for absorption of the body will also vary only with the temperature. The dependence upon the temperature can of course be different for each wave length."

8. This is an interesting observation, which shows that Planck himself was confused.

9. Actually, it is not that hard to check if it is you, or Planck that are confused on the subject.

It all seems to boil down to the question if blackbody radiation is dependent on the surrounding or not.

If you read, and understand sections I and II in the paper "A blackbody is not a blackbox" (Published:http://iopscience.iop.org/0143-0807/32/5/002;the arXiv: http://arxiv.org/pdf/1010.5696v2.pdf), it should be more than obvious that the black body radiation spectra only depends on the temperature of the emitting body.

10. Notice that Planck says "capacity" for absorption/emission is independent of surroundings, which is different from saying that a body absorbs/emits independent of surrounding elements. Of course the "capacity" of a murderer to murder mat be independent of victim, but of course any actual murder depends on the presence and nature of a victim.

11. Planck was not confused when stating "capacity" but appearently this is misunderstood to mean actual emission which always occurs in interplay with surroundings.

12. @Claes Johnson

Planck writes in his Book "Wärmestrahlung" on page 7:

"Doch es gilt allgemein der Erfahrungssatz, dass die Emission eines Körperelements nur von den Vorgängen innerhalb des Körperelements. Ein Körper A von 100 Grad Celsius emittiert gegen einen ihm gegenüber befindlichen Körper B von 0 Grad Celsius genau dieselbe Waermestrahlung, wie gegen einen gleichgroßen und gleichgelegenen Körper B' von 1000 Grad Celsius", ..."

I think it is you who confuses something.

13. It is remarkable that Planck says "es gilt allegemein der Erfahrungssats" which is something like "it is a general experience". It is not clear how to interprete this statement. What is "general experience"? Or does Planck mean "common experience" like feeling heat from a stove? The fact is that only the net heat transfer can be measured and thus really experienced. Planck probably means something like "common assumption" which can be anything.

14. @ Günter Heß,

First there is a distinction between Kirchhoff’s Law of Thermal Emission and Planck’s equation for thermal spectra. Second, a blackbody is defined as one which absorbs all light that falls upon it, i.e. there is no reflection. Third, at thermal equilibrium Emission = Absorption. Fourth, there is a distinction between a blackbody and a dead black. A dead black absorbs all radiation that falls upon it (no reflection) but also emits nothing. An interesting example of a dead black is the wings of a Monarch butterfly.

Kirchhoff’s Law of Thermal Emission: The radiation within an arbitrary cavity at thermal equilibrium is always blackbody radiation, and the ratio between the emissive power E and absorptive power A is a universal function of only temperature T and frequency v: [E/A = f(T, v)]. The nature of the materials plays no part in determining the radiation. Also, Emission = Absorption at thermal equilibrium.

Planck’s equation for thermal spectra: Is universal: independent of the nature of the material emitting thermal radiation, dependent only upon temperature and frequency; in accordance with Kirchhoff’s Law of Thermal Emission.

Notes:
(1) Emission = Absorption at thermal equilibrium is actually Stewart’s Law of Equivalence. Kirchhoff (1859) ‘borrowed’ it from Stewart (1858), without acknowledging Stewart’s priority.